# FIND NUMBER OF TERMS IN ARITHMETIC SERIES GIVEN SUM

Find Number of Terms in Arithmetic Series Given Sum :

To find the first term, common difference, number of terms and sum of an arithmetic series, we use one of the formulas given below.

an  =  a + (n - 1)d

Sn  =  (n/2) [a + l] (or)

Sn  =  (n/2) [2a + (n - 1)d]

## Find Number of Terms in Arithmetic Series Given Sum - Examples

Question 1 :

Given a = 2 , d = 8, Sn = 90 find n and a n

Solution :

S n  =  (n/2) [2a + (n - 1) d]

90  =  (n/2) [2 (2) + (n - 1) 8]

90 x 2  =  n [4 + 8 n - 8]

180  =  n [-4 + 8 n]

180  =  -4 n  + 8 n²

8 n² - 4 n - 180  =  0

Divide the whole equation by 4

2n² - n - 45 = 0

(n - 5) (2 n + 9) = 0

n = 5   or n = -9/2

The number of terms may not be negative or fraction.

a5 =  a + 4 d

a5   =  2 + 4 (8)

a5  =  2 + 32

a5  =  34

Question 2 :

Given a = 8 , an = 62, Sn = 210 find n and d

Solution :

an  =  a + (n - 1) d

an  =  8 + (n - 1) d

62  =  8 + (n - 1) d

62 - 8  =  (n - 1) d

54  =  (n - 1) d -----(1)

Sn  =  (n/2) [2a + (n - 1)d]

210 = (n/2) [ 2 (8) + (n - 1) d]

By applying the value of (n - 1) d in (1), we get

210 = (n/2)[16 + 54]

210 = (n/2) [70]

210/70 = (n/2)

3 x 2 = n

n = 6

(n-1) d = 54

(6-1)d = 54

5d = 54

d = 54/5

Question 3 :

Given an = 4, d = 2, Sn = -14 find n and a

Solution :

an = a + (n - 1) d

an = a + (n - 1) (2)

4 = a + 2 n - 2

4 + 2 = a + 2 n

6 = a + 2 n

a + 2 n = 6 ------(1)

an = l = 4

Sn = (n/2) [a + l]

-14 = (n/2) [ a + 4]

(-14 x 2) = n [ a + 4]

-28 = n[6- 2 n + 4]

-28 = n [10 - 2 n]

-28 = 10 n - 2 n²

2n² - 10 n - 28 = 0

n² -  5 n - 14 = 0

(n - 7) (n + 2) = 0

n = 7  or n = -2

By applying the value of n in (1), we get

a + 2(7) = 6

a + 14 = 6

a = 6 - 14

a = -8

After having gone through the stuff given above, we hope that the students would have understood, find number of terms in arithmetic series given sum.

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